Puzzle:Two players pick consecutively 1,2,3 or 4 stones from a stack of 101 stones. The player who picks the last stone wins. Suppose that both players play perfect, does the first or second player win? What about the situation that the one who picks the last stone loses?
My answer: We first look at the situation where the person that picks the last stone loses. Let player one choose an arbitrary amount of stones and make sure as player two that you pick an amount such that the sum of the picked stones that round is 5 (Note that this is always possible, no matter what player 1 choses). After repeating this process 20 rounds, 100 stones are picked and it's player one's turn. Player 1 can now only choose 1 and pick the last stone, conclusion: player 2 wins.
I am not quite sure what strategy would work for the case that the player who picks the last stone wins. Any help would be appreciated.
Best Answer
Almost the same strategy works: suppose that player $1$'s first move is to pick one stone. Then there are exactly $100$ stones left, and at this point player one cane make use the strategy that you were describing above. Since $100$ is divisible by $5$, player $1$ is sure to win the game.